Precalculus Chapter P.7: Equations – Fundamental Concepts of Algebra

Section P.7: Equations

Objectives

This section covers essential techniques for solving various types of equations encountered in precalculus, including linear, rational, absolute value, quadratic, and radical equations. Mastery of these methods is foundational for further study in mathematics.

• Solve linear equations in one variable, including those with fractions.

• Solve rational equations with variables in the denominator.

• Solve a formula for a variable (rearrange formulas).

• Solve equations involving absolute value.

• Solve quadratic equations by factoring, square root property, completing the square, and using the quadratic formula.

• Use the discriminant to determine the number and type of solutions of quadratic equations.

• Solve radical equations.

Linear Equations

Definition of a Linear Equation

A linear equation in one variable x is an equation that can be written in the form:

$ax + b = 0$

• a and b are real numbers, and $a \neq 0$.

Generating Equivalent Equations

To solve equations, we often transform them into equivalent forms using the following operations:

• Simplify expressions by removing grouping symbols and combining like terms.

• Add or subtract the same real number or variable expression on both sides.

• Multiply or divide both sides by the same nonzero quantity.

• Interchange the two sides of the equation.

Steps for Solving a Linear Equation

1. Simplify each side by removing grouping symbols and combining like terms.

2. Collect all variable terms on one side and all constant terms on the other.

3. Isolate the variable and solve.

4. Check the proposed solution in the original equation.

Example: Solving a Linear Equation

Solve and check: $4(2x+1) = 29 + 3(2x-5)$

1. Simplify both sides: $4(2x+1) = 29 + 3(2x-5)$ $8x + 4 = 29 + 6x - 15$ $8x + 4 = 6x + 14$

2. Collect variable and constant terms: $8x - 6x + 4 = 6x - 6x + 14$ $2x + 4 = 14$ $2x = 10$

3. Isolate the variable: $x = 5$

4. Check: $4(2(5)+1) = 29 + 3(2(5)-5)$ $4(11) = 29 + 3(5)$ $44 = 44$ (True)

Linear Equations Involving Fractions

Solving Equations with Fractions

To solve equations containing fractions, multiply both sides by the least common denominator (LCD) to clear the fractions.

Example:

Solve and check: $\frac{x-3}{4} = \frac{5}{14} - \frac{x+5}{7}$

• LCD is 28; multiply both sides by 28: $28\left(\frac{x-3}{4}\right) = 28\left(\frac{5}{14}\right) - 28\left(\frac{x+5}{7}\right)$ $7(x-3) = 2(5) - 4(x+5)$ $7x - 21 = 10 - 4x - 20$ $7x - 21 = -4x - 10$

• Collect terms and solve: $7x + 4x = -10 + 21$ $11x = 11$ $x = 1$

Rational Equations

Definition and Solution Method

A rational equation contains one or more rational expressions, possibly with variables in the denominator. To solve, clear denominators by multiplying both sides by the least common denominator (LCD).

Example:

Solve: $\frac{5}{x+3} - \frac{6}{x-2} = 0$

• Factor denominators and find LCD: $(x+3)(x-2)$

• Multiply both sides by LCD and solve for $x$.

Solving a Formula for a Variable

Isolating a Variable

To solve a formula for a variable, rearrange the equation so the desired variable is isolated on one side.

• Example: Given $af + pf = pa$, solve for $f$.

• Combine like terms and isolate $f$.

Equations Involving Absolute Value

Definition and Solution Method

The absolute value of $x$ is the distance of $x$ from zero on the number line. To solve $|u| = c$ (where $c > 0$), rewrite as $u = c$ or $u = -c$.

Example:

Solve: $4|1-2x| = 20$

• Divide both sides by 4: $|1-2x| = 5$

• Set up two equations: $1-2x = 5$ or $1-2x = -5$

• Solve for $x$ in each case.

Quadratic Equations

Definition

A quadratic equation in $x$ is an equation of the form:

$ax^2 + bx + c = 0$

• $a$, $b$, and $c$ are real numbers, $a \neq 0$.

Zero-Product Principle

If the product of two expressions is zero, then at least one factor must be zero:

If $AB = 0$, then $A = 0$ or $B = 0$.

Solving by Factoring

1. Rewrite in standard form: $ax^2 + bx + c = 0$.

2. Factor completely.

3. Set each factor equal to zero and solve.

4. Check solutions in the original equation.

Example:

Solve: $2x^2 = 1 - x$

• Move all terms to one side: $2x^2 + x - 1 = 0$

• Factor: $(2x - 1)(x + 1) = 0$

• Set each factor to zero: $2x - 1 = 0$ ($x = \frac{1}{2}$), $x + 1 = 0$ ($x = -1$)

Solving by Square Root Property

For equations of the form $u^2 = d$:

$u^2 = d \implies u = \sqrt{d}$ or $u = -\sqrt{d}$

Or, $u = \pm \sqrt{d}$

Example:

Solve: $3x^2 - 21 = 0$

• $3x^2 = 21$

• $x^2 = 7$

• $x = \pm \sqrt{7}$

Completing the Square

To solve $x^2 + bx = c$ by completing the square:

• Add $\left(\frac{b}{2}\right)^2$ to both sides to form a perfect square trinomial.

• Rewrite as $(x + \frac{b}{2})^2 =$ (new constant).

• Solve for $x$ by taking square roots.

Example:

Solve: $x^2 + 4x - 1 = 0$

• $x^2 + 4x = 1$

• Add $4$to both sides:$x^2 + 4x + 4 = 5$

• $(x + 2)^2 = 5$

• $x + 2 = \pm \sqrt{5}$

• $x = -2 \pm \sqrt{5}$

Quadratic Formula

The solutions of $ax^2 + bx + c = 0$ are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Example:

Solve: $2x^2 + 2x - 1 = 0$

• $a = 2$, $b = 2$, $c = -1$

• $x = \frac{-2 \pm \sqrt{2^2 - 4(2)(-1)}}{2 \cdot 2}$

• $x = \frac{-2 \pm \sqrt{4 + 8}}{4}$

• $x = \frac{-2 \pm \sqrt{12}}{4}$

• $x = \frac{-2 \pm 2\sqrt{3}}{4} = \frac{-1 \pm \sqrt{3}}{2}$

Discriminant

The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is $b^2 - 4ac$. It determines the number and type of solutions:

Discriminant ($b^2 - 4ac$)	Number and Type of Solutions
Positive ($>0$)	Two unequal real solutions
Zero ($=0$)	One real repeated solution
Negative ($<0$)	Two complex (imaginary) solutions

Example:

For $3x^2 - 2x + 5 = 0$:

• $b^2 - 4ac = (-2)^2 - 4(3)(5) = 4 - 60 = -56$

• Since the discriminant is negative, there are two imaginary solutions.

Radical Equations

Definition and Solution Method

A radical equation is one in which the variable appears under a root (square root, cube root, etc.). To solve:

1. Isolate the radical on one side.

2. Raise both sides to the appropriate power to eliminate the radical.

3. Solve the resulting equation. If radicals remain, repeat steps as needed.

4. Check all proposed solutions in the original equation (to avoid extraneous solutions).

Example:

Solve: $\sqrt{x+3} + 3 = x$

• Isolate the radical: $\sqrt{x+3} = x - 3$

• Square both sides: $x + 3 = (x - 3)^2$

• Expand: $x + 3 = x^2 - 6x + 9$

• Rearrange: $0 = x^2 - 7x + 6$

• Factor: $(x - 6)(x - 1) = 0$

• Possible solutions: $x = 6$, $x = 1$

• Check both in the original equation:

  • For $x = 6$: $\sqrt{6+3} + 3 = 6 \implies 3 + 3 = 6$ (True)

  • For $x = 1$: $\sqrt{1+3} + 3 = 1 \implies 2 + 3 = 1$ (False, extraneous)

• Final solution: $x = 6$

*Additional info: Some steps and explanations have been expanded for clarity and completeness.*